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gurpreetsingh
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15 Sep 2010, 07:48
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Time Speed and Distances
I expect after reading this, you guys won't fear from time speed and distance problems.
# Remember the basic formula : Distance = Time x Speed
This formula involves 3 variable and large number of permutations and combination are possible to twist the problem. Lets analyze three main cases :
1. Distance is constant, speed and time are varied.
=> \(Time\) α \(\frac{1}{speed}\) => \(\frac{T_{1}}{T_{2}} = \frac{V_{2}}{V_{1}}\)
Result: If the speed is doubled, the time is halved and if the speed is \(\frac{m}{n}\) 'th of the original speed, the time required is \(\frac{n}{m}\) 'th.
Lets practice this.
Eg1. A person leaves his home everyday at 11:00 am and reaches his office at 12:00 pm. One day he left his house at normal time but traveled the first half of the distance at speed of \(\frac{2}{3}\) of the normal speed. What should be the speed of second half so that he reaches at the same time?
Solution:1
Time taken to reach the office normally = 60 minutes.
=> at half the distance time taken = 30 minutes. If the traveling speed is \(\frac{2}{3}\) of the normal speed for the first half, the time taken is \(\frac{3}{2}\) of the time taken to reach the first half.
=> \(\frac{3}{2}* 30\) = 45 minutes.
To reach the office after 1 hour he needs to travel the second half in 60-45 = 15 minutes.
With the normal speed he travels the second half in 30 minutes, now using the above result if he needs to cover the second half in 15 minutes, he should double his speed.
Eg2. If in the eg1, if he travels the second half at \(\frac{3}{2}\) of the original speed, at what time will he reach the office?
Solution 2:
Using eg1, he will reach the first half in 45 minutes i.e. 11:45.
If he travels the second half at speed = \(\frac{3}{2}\) of normal speed, time required will be
=\(\frac{2}{3}\) of the normal time = \(\frac{2}{3}*30\) = 20 minutes.
Thus he will reach the office at 12:05 pm i.e. 5 minutes late.
Eg3. If 'GG'
travels at the \(\frac{3}{4}\) normal speed, he is late by 15 minutes. How much time usually he takes to reach the office?
Solution 3:
New speed = \(\frac{3}{4}\) of original
New time taken = \(\frac{4}{3}\) of original
Difference in time = new time - old time = \(\frac{4t}{3} - t = \frac{t}{3}= 15\) minutes
=> t = 45 minutes
Lets crack a tough problem.
Eg4. A, B, and C starts from the same place and travel in the same direction at speeds of 30,40,60 respectively.
B starts 2 hours after A, but B and C overtakes A at the same instant. How many hours after A did C start?
Looks daunting? Don't worry. Lets crack it.
Solution 4:
Since the distance traveled by each of them is same, we can use the concept we had already discussed.
Time taken by A = T , speed of A = 30
Time taken by B = T -2 , speed of A = 40
Time taken by C = T - c , speed of A = 60
Now we have , \(\frac{T}{(T-2)} = \frac{40}{30}\)
=> T = 8
Also we have, \(\frac{T}{(T-c)} = \frac{60}{30}\) , put T=8
we get c=4
=> C started 4 hours after A.
2. Time is constant, speed and Distance are varied.
\(Distance\) α \(speed\) => \(\frac{D_{1}}{D_{2}} = \frac{V_{1}}{V_{2}}\)
Result : If the speed of travel is doubled, the distance traveled in the same time is doubled.
If the speed is \(\frac{m}{n}\) 'th of the original speed, the Distance traveled in the same time is \(\frac{m}{n}\) 'th. of the original distance.
Lets practice this.
Eg5. X and Y run a race between A and B stations, 5 Kms apart. X starts at 9 AM from A at speed of 5 km/h, reaches B and returns back to A at same speed. Y starts at 9:45 AM from A at speed 10 km/h, reaches B and comes back to A at same speed.
At what time do X and Y first meet each other?
Solution 5: Since distance is 5 Kms, X reaches B at 10:00 AM. In 15 minutes (at 10:00 AM) Y has traveled 2.5 kms i.e. half the distance between stations.
Now X is traveling towards A and Y towards B. From 10:00 to the time till they reach they travel for the same time.
=> Ratio of their speed = Ratio of distance traveled by them
=> \(10/5 = \frac{D_{Y}}{D_{X}}\)
=> \(2 * D_{X} = D_{Y}\) => Distance traveled by Y = twice that of X.
=> Distance traveled by Y = \(\frac{2}{3} *\) half of AB = \(\frac{2}{3} * \frac{5}{2}\) = 5/3
Time taken = \(\frac{5}{3} * \frac{60}{10}= 10\) minutes => they will meet at 10:10 Am.
Will be updated soon with more Time speed and distance related fundamentals, keep visiting the thread
I expect after reading this, you guys won't fear from time speed and distance problems.
# Remember the basic formula : Distance = Time x Speed
This formula involves 3 variable and large number of permutations and combination are possible to twist the problem. Lets analyze three main cases :
1. Distance is constant, speed and time are varied.
=> \(Time\) α \(\frac{1}{speed}\) => \(\frac{T_{1}}{T_{2}} = \frac{V_{2}}{V_{1}}\)
Result: If the speed is doubled, the time is halved and if the speed is \(\frac{m}{n}\) 'th of the original speed, the time required is \(\frac{n}{m}\) 'th.
Lets practice this.
Eg1. A person leaves his home everyday at 11:00 am and reaches his office at 12:00 pm. One day he left his house at normal time but traveled the first half of the distance at speed of \(\frac{2}{3}\) of the normal speed. What should be the speed of second half so that he reaches at the same time?
Solution:1
Time taken to reach the office normally = 60 minutes.
=> at half the distance time taken = 30 minutes. If the traveling speed is \(\frac{2}{3}\) of the normal speed for the first half, the time taken is \(\frac{3}{2}\) of the time taken to reach the first half.
=> \(\frac{3}{2}* 30\) = 45 minutes.
To reach the office after 1 hour he needs to travel the second half in 60-45 = 15 minutes.
With the normal speed he travels the second half in 30 minutes, now using the above result if he needs to cover the second half in 15 minutes, he should double his speed.
Eg2. If in the eg1, if he travels the second half at \(\frac{3}{2}\) of the original speed, at what time will he reach the office?
Solution 2:
Using eg1, he will reach the first half in 45 minutes i.e. 11:45.
If he travels the second half at speed = \(\frac{3}{2}\) of normal speed, time required will be
=\(\frac{2}{3}\) of the normal time = \(\frac{2}{3}*30\) = 20 minutes.
Thus he will reach the office at 12:05 pm i.e. 5 minutes late.
Eg3. If 'GG'
travels at the \(\frac{3}{4}\) normal speed, he is late by 15 minutes. How much time usually he takes to reach the office? Solution 3:
New speed = \(\frac{3}{4}\) of original
New time taken = \(\frac{4}{3}\) of original
Difference in time = new time - old time = \(\frac{4t}{3} - t = \frac{t}{3}= 15\) minutes
=> t = 45 minutes
Lets crack a tough problem.
Eg4. A, B, and C starts from the same place and travel in the same direction at speeds of 30,40,60 respectively.
B starts 2 hours after A, but B and C overtakes A at the same instant. How many hours after A did C start?
Looks daunting? Don't worry. Lets crack it.
Solution 4:
Since the distance traveled by each of them is same, we can use the concept we had already discussed.
Time taken by A = T , speed of A = 30
Time taken by B = T -2 , speed of A = 40
Time taken by C = T - c , speed of A = 60
Now we have , \(\frac{T}{(T-2)} = \frac{40}{30}\)
=> T = 8
Also we have, \(\frac{T}{(T-c)} = \frac{60}{30}\) , put T=8
we get c=4
=> C started 4 hours after A.
2. Time is constant, speed and Distance are varied.
\(Distance\) α \(speed\) => \(\frac{D_{1}}{D_{2}} = \frac{V_{1}}{V_{2}}\)
Result : If the speed of travel is doubled, the distance traveled in the same time is doubled.
If the speed is \(\frac{m}{n}\) 'th of the original speed, the Distance traveled in the same time is \(\frac{m}{n}\) 'th. of the original distance.
Lets practice this.
Eg5. X and Y run a race between A and B stations, 5 Kms apart. X starts at 9 AM from A at speed of 5 km/h, reaches B and returns back to A at same speed. Y starts at 9:45 AM from A at speed 10 km/h, reaches B and comes back to A at same speed.
At what time do X and Y first meet each other?
Solution 5: Since distance is 5 Kms, X reaches B at 10:00 AM. In 15 minutes (at 10:00 AM) Y has traveled 2.5 kms i.e. half the distance between stations.
Now X is traveling towards A and Y towards B. From 10:00 to the time till they reach they travel for the same time.
=> Ratio of their speed = Ratio of distance traveled by them
=> \(10/5 = \frac{D_{Y}}{D_{X}}\)
=> \(2 * D_{X} = D_{Y}\) => Distance traveled by Y = twice that of X.
=> Distance traveled by Y = \(\frac{2}{3} *\) half of AB = \(\frac{2}{3} * \frac{5}{2}\) = 5/3
Time taken = \(\frac{5}{3} * \frac{60}{10}= 10\) minutes => they will meet at 10:10 Am.
Will be updated soon with more Time speed and distance related fundamentals, keep visiting the thread
15 Sep 2010, 11:26
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good work keep going............
